First slide

Applications of differential equations

Question

The length of latus rectum of the conic passing through the origin and having the property that normal at each point (x, y) intersects the x-axis at $((x + 1), 0)$ is:

Moderate

A

1
B

2
C

4
D

none of these

Solution

Let $p (x, y)$ be any point on the conic.
$∴$ slope of normal $l = - \frac{d x}{d y}$
Equation of normal to conic at $p (x, y)$ is
$(Y - y) = - \frac{d x}{d y} (X - x)$
$\Rightarrow - y = - \frac{d x}{d y} (X - x)$ (on x-axis Y = 0)
$\Rightarrow X = x + y \frac{d y}{d x}$
Given $X = x + 1 \Rightarrow y \frac{d y}{d x} = 1$
$y d y = d x$
$\Rightarrow \frac{y^{2}}{2} = x + C . . . . . . . (i)$
Since the conic passes through origin, x = 0, y = 0 in (i) we get C = 0
$∴$ From $(i) y^{2} = 2 x$
Which is parabola having length of latus rectum = $4 a = 2 u n i t s$

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